counting triangles

Suppose that each side of an equilateral triangle can be partitioned into n partitions, all of which have equal length. From this, you are able to create n^2 equilateral triangles. These are n triangles. From this information, how many DOWNWARD pointing triangles are there in an n triangle?

Suppose that each side of an equilateral triangle can be partitioned into n partitions, all of which have equal length. From this, you are able to create n^2 equilateral triangles. These are n triangles. From this information, how many DOWNWARD pointing triangles are there in an n triangle?

Apparently its a messy formula.

Any thoughts?

Lets assume we are talking about small downward pointing equilateral
triangles.

There are n-rows of triangles.

The first contains 0 down pointers,
the second row contains 1 down pointer,
the third row contains 2 down pointer,
:
:
the i-th row contains i-1 down pointer,
:
:
the n-th row contains n-1 down pointer.

Lets assume we are talking about small downward pointing equilateral
triangles.

There are n-rows of triangles.

The first contains 0 down pointers,
the second row contains 1 down pointer,
the third row contains 2 down pointer,
:
:
the i-th row contains i-1 down pointer,
:
:
the n-th row contains n-1 down pointer.

So there are:

RonL

Thanks for your help Ron, but I'm afraid it's not that easy. I have to count ALL the triangles, and thus it's every possible triangle, not just the small ones. That's what makes this problem very difficult.

Thanks for your help Ron, but I'm afraid it's not that easy. I have to count ALL the triangles, and thus it's every possible triangle, not just the small ones. That's what makes this problem very difficult.

The usual form of the questions would normaly refer to triangles of any size,
but the wording of this one suggested it was only interested in the small
ones.

So do you know how many downward pointing triangles there including the big triangles.

For example, suppose we have 1 triangle, then

y(1) = 0

Then partitioned into 2 rows..

y(2) = 1

.
.
.
y(4) = 7

.
.
.
etc

Hmm, okay, so the pattern looks like follows:

y(1) = 0
y(2) = 1
y(3) = 3
y(4) = 7
y(5) = 13
y(6) = 21
y(7) = 31

.
.
.

Now it's obvious the pattern is just adding an extra 2 on the prev. sum. Thus, y(8) will be 31 + 12 = 43, since the prev term was 13 + 10.

What's frustrating me is that I can't come up with a formula to express this, and this should be the easy part. The part that should be difficult for this is proving why it's true! It's obvious looking at the pictures but I have to PROVE it. My hint was to use difference equations.